Problem. Fix a prime $p$ and a positive integer $n$. For each $j$ put $$\Gamma_{j} = \{g \in \mathrm{SL}_{n}(\mathbb{Z}_{p}) \mid g \equiv 1_{n} \text{ mod }p^{j}\}.$$

(i) Show that $\Gamma_{1}$ is a pro-$p$ group (with topology induced by $p$-adic topology on $\mathrm{M}_{n}(\mathbb{Z}_{p})$).

(ii) Show that $\Gamma_{1}$ is topologically finitely generated. Deduce that every subgroup of finite index in $\mathrm{SL}_{n}(\mathbb{Z}_{p})$ contains $\Gamma_{j}$ for some $j$.

My attempt.

(i) I have a little doubt of what to use to solve the result. My teacher told me, first, to prove that $\Gamma_{1}$ is a closed subgroup of $\mathrm{M}_{n}(\mathbb{Z}_{p})$.

I know that $\mathrm{M}_{n}(\mathbb{Z}_{p})$ is compact, Hausdorff and totally disconnected and, for each $j$, $\Gamma_{j} = \ker(\pi_{j})$ where $\pi_{j}: \mathrm{SL}_{n}(\mathbb{Z}_{p}) \to \mathrm{SL}_{n}(\mathbb{Z}/p^{j}\mathbb{Z})$, then, $\Gamma_{j} \triangleleft_{c} \mathrm{M}_{n}(\mathbb{Z}_{p})$.

But I don't know how to use it. The only result about pro-$p$ groups and closed subgroups are (assuming $G$ profinite)

  1. If $G$ is pro-$p$ and $H \leq_{c} G$, then $H$ is pro-$p$.

  2. Let $K \triangleleft_{c} G$. Then $G$ is pro-$p$ iff both $K$ and $G/K$ are pro-$p$ groups.

But both results don't seem useful. Also, I tried (I'm not sure if it's true):

$$\mathrm{SL}_{n}(\mathbb{Z}_{p}) \simeq \varprojlim \mathrm{SL}_{n}(\mathbb{Z}/p^{j}\mathbb{Z}) \simeq \varprojlim \mathrm{SL}_{n}(\mathbb{Z})/\Gamma_{j},$$ then, $$\Gamma_{1} = \varprojlim \Gamma_{j}.$$ So, if I prove that if $\Gamma_{j}$ is a $p$-group, the problem is solved. But I don't know to show it.

(ii) Is possible to prove that $\Gamma_{j}$ is a base of the neighborhoods of $1_{n}$? I'm not sure if it's is true, but if yes, since $\Phi(\Gamma_{1}) = \Gamma_{2}$, $\Gamma_{1}$ is finitely generated. Also, if $H$ is a subgroup of finite index in $\mathrm{SL}_{n}(\mathbb{Z}_{p})$, then $H \cap \Gamma_{1}$ is an open subgroup of finite index of $\Gamma_{1}$ so, contains $\Gamma_{j}$ for some $j$.

Thanks for the advance.

  • $\begingroup$ (i) $M_n(Z_p)$ is only a monoid under multiplication. So, well, indeed it's a closed subgroup of this monoid, but it sounds unusual a teacher terms things in this way. $\endgroup$
    – YCor
    May 27, 2019 at 18:09
  • $\begingroup$ @YCor In fact, my second attempt seems more natural for me. $\endgroup$
    – Lucas
    May 27, 2019 at 18:13

1 Answer 1


(i) By definition, $$ \begin{aligned} \Gamma_{j} &= \{\ g \in \operatorname{SL}_{n}(\mathbb{Z}_{p}) \ \mid\ g \equiv 1_{n} \text{ mod }p^{j}\ \} \\ &= \left\{\ A=\left[\begin{smallmatrix}a&b&\dots\\c & d &\dots\\ \vdots &\vdots &\ddots\end{smallmatrix} \right] \in \operatorname{M}_{n\times n}(\mathbb{Z}_{p}) \ \mid\ \det A=1\ , |a-1|,\ |b-0|,\ |c-0|, |d-1|,\dots\le p^{-j}\ \right\} \ , \end{aligned} $$ and the conditions imposed to $a,b,c,d,\dots$ are preimages through continuous functions of closed subsets, so we obtain a closed subset.

To see it is a pro-$p$-group, let us compute the order of $\Gamma_1/\Gamma_j$. This is isomorphic to the subgroup of all matrices with entries $a,b,c,d,\dots\in\Bbb Z/p^j$ with $a-1,b-0,c-0,d-1\dots\in p^1\Bbb Z$. We let $b,c,d,\dots$ take arbitrary values, obtain a number of possibilities $=$ power of $p$, and this determines $a$ from the determinant one condition, since the diagonal elements $d,\dots$ is/are invertible, and of course $a=1$ modulo $p$. The groups $\Gamma_j$ are building a basis of neighborhoods of the neutral element (see (ii)), so each topological (normal) open subgroup is included in such a $\Gamma_j$.

(ii) Which is the topology of the normed space $V=M_{n\times n}(\Bbb Z_p)$ of all $n\times n$ matrices over $\Bbb Z_p$? It is the product topology of the $n^2$ copies of of $F=\Bbb Z_p$ used for the entries, i.e. topologically $V\cong F^{n\times n}$.

A basis of neighborhoods of a point / a matrix $A\in V$ reduces by translation (since addition of $A$ / subtraction by $A$ are continuous, inverse to each other) to a basis of neighborhoods of the zero matrix $0$, with $n^2$ zero entries. Then the "boxes" of the shape

$B(j(a))=p^{j(a)}\Bbb Z_p$,

$B(j(b))=p^{j(b)}\Bbb Z_p$,

$B(j(c))=p^{j(c)}\Bbb Z_p$,

$B(j(d))=p^{j(d)}\Bbb Z_p$,

are neighborhoods of the zero in the entry $a,b,c,d,\dots$ respectively.

The cartesian product box $B(j(a))\times B(j(b))\times B(j(c))\times B(j(d))\times\dots$ is then a typical element in the product basis. The boxes of the shape $B(j)\times B(j)\times B(j)\times B(j)\times\dots$ are filtering.

Shifting this zero-neighborhoods of $V$ to the identity matrix, and intersecting with $\Gamma_1$ we obtain the $\Gamma_j$ sets, by definition of the restricted topology they form a base for the topology of $\Gamma_1$.

To see that we have a topologically finite generation of $\Gamma_1$ and/or of $\operatorname{SL}_n(\Bbb Z_p)$, we need some decomposition properties. An LU-decomposition reduces to the case of lower / upper triangular matrices, then we arrange step by step (diagonally) using a topological generator $pt_{ij}$ for each non-diagonal position for the additive group $(p\Bbb Z_p, +)$, the using the matrix $1+t_{ij}E_{ij}$ with the elementary matrix $E_{ij}$. (For $\operatorname{SL}_n(\Bbb Z_p)$ one can argue that $\operatorname {SL}_n(\Bbb Z)$ sits dense in it, then apply a fraction-free LU-decomposition to get (true group) generators for $\operatorname {SL}_n(\Bbb Z)$. Then we are in a known position, link. Well, this approximation with integers is not immediately trivial, but we can approximate (with the same idea as above) all but one entry (say that diagonal $a$ at position $(1,1)$ in a matrix $A$) with elements in $\Bbb Z$, then this last entry is one modulo $p$, so we can find an integer approximating it. This argument was designed for the full $\operatorname {SL}_n(\Bbb Z_p)$ group, analogous arguments should work for $\Gamma_1$ and $\Gamma_j$).

Somehow i have the feeling that i am missing the point of the quesiton, since from the precise high level of the question you should know all of the above topological ingredients better, they are already shipped in the package.

  • $\begingroup$ Great answer! I'll check the details, but this answer seems extremely clear to me (indeed, it's even more explained than I needed). Thank you very much! $\endgroup$
    – Lucas
    May 29, 2019 at 3:15
  • $\begingroup$ Nice +1! Can you explain the isomorphism you said? I can't see why $a,b,c,d... \in \Bbb{Z }/p^j$. $\endgroup$
    – Greg
    Apr 17 at 22:15
  • 1
    $\begingroup$ @Greg I'll work with $n=2$ for an easy typing. Consider the map$$\Gamma_1\to\operatorname{GL}_2(\Bbb Z/p^j)\ ,$$ which maps a matrix with entries $a,b,c,d\in\Bbb Z_p$ to the matrix with entries $a$ mod $p^kj$, $b$ mod $p^kj$, $c$ mod $p^kj$, $d$ mod $p^kj$. (Here we identify $\Bbb Z_p/p^j$ with $\Bbb Z/p^j$, induced by the inclusion $\Bbb Z\to\Bbb Z_p$.) Its image ist not the full GL${}_2$, but the part $H$ of it of all matrices with entries $a',b',c',d'$ with $a-1, b-0, c-0, d-1\in p\Bbb Z/p^j\Bbb Z$, the group mentioned in the answer, part (i). Its kernel is $\Gamma_j$. $\endgroup$
    – dan_fulea
    Apr 19 at 10:40
  • 1
    $\begingroup$ ... So we have an exact sequence:$$1\to \Gamma_j\to \Gamma_1\to H\to 1\ .$$For a general $n$ the same argument applies. Please do not hesitate to ask, here or else, if there is some unclear point! @Greg $\endgroup$
    – dan_fulea
    Apr 19 at 10:42
  • $\begingroup$ I got the idea. I have just one question. Why do you define $a$ mapped to $a \pmod{p^kj}$? The idea is embedded the entries of a matriz $A$ into a matriz with entries in $\Bbb{Z}/p^j\Bbb{Z}$ (or $\Bbb{Z}/p^j\mathbb{Z}$), right? $\endgroup$
    – Greg
    Apr 19 at 17:25

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